Appendix D Selected Solutions - d3dcoder.net · Solution: a) ... of Exercise 14, we can conclude: ... Appendix D Selected Solutions, Introduction to 3D Game Programming with DirectX

Appendix D Selected Solutions, Introduction to 3D Game Programming with DirectX 11, by

Frank D. Luna

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Appendix D Selected Solutions

Chapter 1

1. Let and . Perform the following computations and draw the vectors relative to a 2D coordinate system.

a) b)

c)

d) Solution:

a)

b)

c)

d) 3. This exercise shows that vector algebra shares many of the nice properties of real numbers (this is not an exhaustive list). Assume , , and

. Also assume that and are scalars. Prove the following vector

properties. a) (Commutative Property of Addition) b) (Associative Property of Addition) c) (Associative Property of Scalar Multiplication) d) (Distributive Property 1) e) (Distributive Property 2) Solution:

a)

b)


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c)

d)

e)

5. Let and . Normalize and . Solution:


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7. Is the angle between and orthogonal, acute, or obtuse? a) , b) , c) ,

a) b) c)

9. Let , , and . Also let and be scalars.

Prove the following dot product properties. a) b) c) d) e) Solution:


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11. Let . Decompose the vector into the sum of two orthogonal vectors, one parallel to and the other orthogonal to . Also, draw the vectors relative to a 2D coordinate system. Solution:

13. Let the following points define a triangle relative to some coordinate system: , , and . Find a vector orthogonal to this triangle. Hint: Find two vectors on two of the triangle’s edges and use the cross product. Solution:

15. Prove that gives the area of the parallelogram spanned by and ; see Figure below.

[Figure D.1: Parallelogram spanned by two 3D vectors and ; the parallelogram has base

and height .] Solution:


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The area is the base times the height:

Using trigonometry, the height is given by . This, along with the application of Exercise 14, we can conclude:

17. Prove that the cross product of two nonzero parallel vectors results in the null vector; that is, . Hint: Just use the cross product definition. Solution:

Chapter 2

1. Solve the following matrix equation for :

.

Solution:

3. Compute the transpose of the following matrices:

a) , b)

, c)


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Solution:

5. Show that

Solution: Let be an arbitrary row in . By definition of matrix multiplication, we know that the ith row is given by

However, by definition of matrix multiplication, this is equal to the vector-matrix product . That is,

Since was an arbitrary row, we just substitute to complete the proof:

7. Prove that the cross product can be expressed by the matrix product:

Solution:


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9. Let

. Is

the inverse of ?

Solution:

Since we can conclude that is not the inverse of .

11. Find the inverse of the following matrices:

Solution:

For the matrix, we have the formula:

To verify:

For the matrix, we use the formula:


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The cofactor matrix is:

So

and

To verify:

13. Show that , assuming is invertible. Solution:

Therefore, is the inverse of .

15. Prove that the 2D determinant

gives the signed area of the parallelogram spanned

by and . The result is positive if can be rotated counterclockwise to

coincide with by an angle , and negative otherwise.


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[Figure D.2: Parallelogram spanned by two 2D vectors and ; the parallelogram has base

and height .]

Solution:

The area is given by base times height:

Squaring both sides:

Taking the square root:

17. Let

,

, and

. Show that .

This shows that matrix multiplication is associative for matrices. (In fact, matrix multiplication is associative for general sized matrices, whenever the multiplication is defined.) Solution: For matrices, we will just do the computations:


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Comparing the terms element-by-element, we see

Chapter 3

1. Let be defined by . Is a linear transformation? If it

is, find its standard matrix representation.

Solution:

If is linear, then we must have:

1. 2.

Let and .

So ; therefore is not a linear transformation.

3. Assume that is a linear transformation. Further suppose that , , and . Find .


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Solution:

We are given that is a linear transformation and its behavior on the standard basis vectors

. Therefore, the standard matrix representation of is:

Then

5. Build a rotation matrix that rotates along the axis .

Solution:


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7. Build a single transformation matrix that first scales 2 units on the x-axis, -3 units on the y-

axis, and keeps the z-dimension unchanged, and then translates 4 units on the x-axis, no units on

the y-axis, and -9 units on the z-axis.

Solution:

Then the desired transformation matrix is obtained by the product

9. Redo Example 3.2, but this time scale the square 1.5 units on the x-axis, 0.75 units on the y-

axis, and leave the z-axis unchanged. Graph the geometry before and after the transformation to

confirm your work.

Solution:

The corresponding scaling matrix is:

Now to actually scale (transform) the square, we multiply both the minimum point and

maximum point by this matrix:


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[Figure D.3: Scaling transform.]

11. Redo Example 3.4, but this time translate the square -5 units on the x-axis, -3.0 units on the

y-axis, and 4.0 units on the z-axis. Graph the geometry before and after the transformation to

confirm your work.

Solution:

The corresponding translation matrix is:

Now to actually translate (transform) the square, we multiply both the minimum point and

maximum point by this matrix:


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[Figure D.4: Translation transform.]

13. Prove that the rows of are orthonormal. For a more computational intensive exercise, the

reader can do this for the general rotation matrix (rotation matrix about an arbitrary axis), too.

Solution:

Let , , and .

First we show all the rows are unit length:

Next we show the rows are mutually orthogonal:

14. Prove the matrix is orthogonal if and only if .

Solution:


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For concreteness, we will prove for matrices since 3D rotation matrices are the only kind

of orthogonal matrices we care about in this book. However, the same argument we present

generalizes to matrices.

Suppose is a orthogonal matrix:

Then, by the definition of matrix multiplication . But, because is orthogonal

we have that

It is true that only for the diagonal elements of the matrix. Thus the resulting matrix has

1’s along the diagonal and zeros everywhere else. But this is exactly the identity matrix. Hence,

A similar argument shows . Therefore, we must have that .

Now suppose that . In particular, this implies that . In turn, this means:

Therefore, the row vectors or are mutually orthogonal and unit length; thus, is orthogonal.

15. Compute:

and

Does the translation translate points? Does the translation translate vectors? Why does it not

make sense to translate the coordinates of a vector in standard position?

Solution:

The translation translates points.


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The translation does not translate vectors. Translation does not make sense for vectors because a

vector only describes direction and magnitude, independent of location.

17. Suppose that we have frames A and B. Let and represent a

point and force, respectively, relative to frame A. Moreover, let ,

,

, and describe frame A with coordinates relative to

frame B. Build the change of coordinate matrix that maps frame A coordinates into frame B

coordinates, and find and . Draw a picture on graph paper to verify

that your answer is reasonable.

Solution:

From Equation 3.9, the change of coordinate matrix is:

Then to transform points and vectors from frame A into frame B we multiply the frame A coordinate

vectors by the matrix:


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[Figure D.5: Change of coordinates from frame A to frame B.]

19. Consider the triangle defined by the points , , and . Graph the following points:

a)

b)

c)

d)

e)

f)

What is special about the point in part (a)? What would be the barycentric coordinates of and

the point in terms of ? Can you make a conjecturer about where the point

will be located relative to the triangle if one of the barycentric coordinates is negative?

Solution:

a)

b)


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c)

d)

e)

f)

The point in part (a) is the centroid. so the barycentric coordinates are

.

so the barycentric coordinates are

. For negative

barycentric coordinates, the points lie outside the triangle.

[Figure D.6: Plotting barycentric coordinates.]

21. Consider Figure 3.16. A common change of coordinate transformation in computer graphics

is to map coordinates from frame A (the square ) to frame B (the square where the

y-axes aims opposite to the one in Frame A). Prove that the change of coordinate transformation

from Frame A to Frame B is given by:


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[Figure D.7: Change of coordinates from frame A (the square ) to frame B (the square

where the y-axes aims opposite to the one in Frame A)]

Solution:

From Equation 3.9, the change of coordinate matrix is:

That is, we need to describe the coordinate system of frame A (origin and axes) with coordinates relative

to frame B. (We use homogeneous coordinates.) The origin in frame A is the center of the square, which

has coordinates . The x-axis of frame A aims in the same direction as frame B, and,

along this axis, one unit in frame A is half a unit in frame B; therefore, . The y-axis of

frame A aims in the opposite direction as frame B, and along this axis, one unit in frame A is half a unit in

frame B; therefore, . Since this is a 2D problem, we will just assume that the z-axes

are the same in both coordinate systems; therefore, . Substituting these numbers into the

above yields the desired transformation matrix:

23. Consider the transformation that warps a square into a parallelogram given by:

Find the standard matrix representation of this transformation, and show that the determinant of

the transformation matrix is equal to the area of the parallelogram spanned by and .


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[FigD.8.bmp: Transformation that maps square into parallelogram.]

Solution:

The area of the parallelogram spanned by by and is given by:

The cross product is not defined in 2D, but we can do a trick and augment to 3D space with

so that we can use the cross product. Note that using does not affect the magnitude

or direction of the vectors.

Also see Chapter 2, Exercise 15. So observe that the area of the parallelogram (square) defined

by the standard basis vectors and is 1, and the area of the parallelogram spanned by and

is 5. Therefore, the transformation changed the volume (or area in 2D) from 1 to 5 when it

warped it from a unit square to the parallelogram spanned by and .

25. A rotation matrix can be characterized algebraically as an orthogonal matrix with

determinant equal to 1. If we reexamine Figure 3.7 along with Exercise 24 this makes sense; the

rotated basis vectors , , and are unit length and mutually orthogonal; moreover,

rotation does not change the size of the object, so the determinant should be 1. Show that the


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product of two rotation matrices is a rotation matrix. That is, show (to show is orthogonal), and show .

Solution:

Using the fact that :

27. Find a scaling, rotation, and translation matrix whose product transforms the line segment

with start point and endpoint into the line segment with length 2,

parallel to the vector , with start point .

Solution:

The line segment currently aims along the z-axis with a length of 1. To make it have a length of

2, we first scale it 2-units on the z-axis. In order to rotate the line segment so that it is parallel to

we need to rotate by an angle in the plane that contains the vectors and

(see Figure). The axis of rotation is given by . The angle we need to rotate is given by the angle between the vectors and :

Finally, we must apply a translation .

[FigD.9: To rotate so that it aims in the same direction as , we must rotate by an angle

about the axis .]

Chapter 5

1. Construct the vertex and index list of a pyramid, as shown in Figure 5.35.


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Solution:

[FigD.10: Vertices of the pyramid.]

Vertex vertices[5] = {v0, v1, v2, v3, v4};

UINT indices[] = {

0, 1, 2,

1, 3, 2,

1, 4, 0,

0, 4, 2,

2, 4, 3,

3, 4, 1

};

3. Relative to the world coordinate system, suppose that the camera is positioned at

and looking at the point . Compute the view matrix assuming describes the “up” direction in the world.

Solution:

Figure 5.20 shows the setup. Let , and . The

direction the camera is looking is given by:

This vector describes the local z-axis of the camera. A unit vector that aims to the “right” of is

given by:

This vector describes the local x-axis of the camera. Finally, a vector that describes the local y-

axis of the camera is given by:


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The view matrix is:

You can verify the answer by writing the following code and inspecting the matrix elements:

XMMATRIX K = XMMatrixLookAtLH(

XMLoadFloat3(&XMFLOAT3(-20.0f, 35.0f, -50.0f)),

XMLoadFloat3(&XMFLOAT3(10.0f, 0.0f, 30.0f)),

XMLoadFloat3(&XMFLOAT3(0.0f, 1.0f, 0.0f)));

5. Suppose that the view window has height . Find the distance from the origin the view

window must be to create a vertical field of view angle .

Solution:

7. Suppose that you are given the following perspective projection matrix with fixed :

Find the vertical field of view angle the aspect ratio , and the near and far plane values that

were used to build this matrix in terms of . That is, solve the following equations:


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Solving these equations will give you formulas for extracting the vertical field of view angle

the aspect ratio , and the near and far plane values from any perspective projection matrix of the

kind described in this book.

Solution:

Dividing the second equation by the first equation yields:

The second equation implies:

We solve the third equation for :

Now we plug into the fourth equation and solve for :

Now,

In summary,


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8. For projective texturing algorithms, we multiply an affine transformation matrix after the

projection matrix. Prove that it does not matter if we do the perspective divide before or after

multiplying by . Let, be a 4D vector, be a projection matrix, be a affine

transformation matrix, and let a subscript denote the -coordinate of a 4D vector, prove:

Solution:

Using the properties of matrix algebra, we have:

But is given as an affine transformation matrix, which means it does not modify the w-

coordinate. Hence, we have . Hence,

10. Let be the coordinates of a point in view space, and let be the

coordinates of the same point in NDC space. Prove that you can transform from NDC space to

view space in the following way:

Explain why you need the division by . Would you need the division by if you were

transforming from homogeneous clip space to view space?

Solution:

Recall from Equation 5.1 and §5.6.3.5 that:


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where are view space coordinates.

The steps to go from view space to NDC space are multiplying by the projection matrix,

followed by a division by . Therefore, to transform back from NDC space to view space,

we multiply by the inverse projection matrix followed by a division by (which is

equivalent to multiplying by ).

If you are given points in homogeneous clip space (before the divide by ), then you

do not need to divide by in the inverse transformation. In homogeneous clip space, the point has coordinates :


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13. Consider the 3D shear transform given by . This

transformation is illustrated in Figure 5.37. Prove that this is a linear transformation and has the

following matrix representation:

[FigD.11: The x- and y-coordinates sheared by the z-coordinate. The top face of the box lies in

the plane. Observe that the shear transform translates points in this plane.]

Solution:


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So is a linear transformation. To find the matrix representation, we apply it to the standard

basis vectors:

The matrix representation is found by inserting the above three vectors into the rows of a matrix:

Chapter 8

6. Let , , and be the vertices of a 3D triangle with respective texture coordinates , ,

and . Recall from §8.2 that for an arbitrary point on a 3D triangle where , its texture coordinates are found

by linearly interpolating the vertex texture coordinates across the 3D triangle by the same parameters:

a) Given and , , and , solve for in terms of and (Hint: Consider the vector

equation . b) Express as a function of and ; that is, find a formula . c) Compute and and give a geometric interpretation of what these vectors mean.

Solution:

a) Let , , and :

From Example 2.10, the inverse of a matrix

is given by:

We can solve for by multiplying by the inverse:


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In other words, given the texture coordinates of a point on a triangle, we can solve for the

parametric coordinates that yield the 3D point corresponding to the texture point .

b) Part (a) showed that and . Therefore: can be expressed in terms of :

c) First, note that:

Now the partial derivatives are given by:


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where we use the notation:

Similarly,

In summary,

Compare this result to the one derived in §18.3; we have simple derived the forumla in §18.3 in a

different way.

The 3D vector

gives us the “velocity” we move in 3D space when we move in the u-

direction in texture space. Put another way, it tells us the direction of the texture space u-axis in

3D space. Likewise, the 3D vector

gives us the “velocity” we move in 3D space when we


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move in the v-direction in texture space (i.e., the direction of the texture space v-axis in 3D

space).

7. See “TexturedColumns” on the DVD.

Chapter 10

2. Prove that

by doing the matrix multiplication for each

component, as was done in §10.5.1 for directional lights.

Solution:

Let . For , the ith coordinate of is given by:

and the fourth coordinate is given by:

Doing the homogeneous divide we obtain:

But this is exactly the ith coordinate of

, so

7. See “DepthStencilBolt” on the DVD.

Chapter 11

2. See “GSSubdivision” on the DVD.


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Chapter 12

5. See “CSGpuWavesDemo” on the DVD.

Chapter 14

1. Given the world space axes and origin in world coordinates: , , and , and the view space axes and origin in world coordinates:

, , and , derive the view matrix

form

using the dot product. (Remember, to find the change of coordinate matrix from world space to

view space, you just need to describe the world space axes and origin with coordinates relative to

view space. Then these coordinates become the rows of the view matrix.)

Solution:

[Figure D.12: Finding the world space coordinates in view space.]

All the given vectors have coordinates in world space, and all the axis vectors are unit vectors.

However, using the dot product, we can obtain view space coordinates of the world space. From

the figure, we see:


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Augmenting to homogeneous coordinates and putting these vectors into the rows of a matrix

yields the view matrix.

Chapter 15

2. The plane equations in NDC space take on a very simple form. All points inside the view

frustum are bounded as follows:

In particular, the left plane equation is given by and the right plane equation is given by

in NDC space. In homogeneous clip space before the perspective divide, all points inside

the view frustum are bounded as follows:

Here, the left plane is defined by and the right plane is defined by . Let

be the view-projection matrix product, and let be a point in world space

inside the frustum. Consider to

show that the inward facing frustum planes in world space are given by:

Left

Right

Bottom

Top

Near

Far

[Notes]

a) We ask for inward facing normals. That means a point inside the frustum has a positive

distance from the plane; in other words, for a point inside the frustum.

b) Note that , so the above dot product formulas do yield plane equations of the form

.


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c) The calculated plane normal vectors are not unit length; see Appendix C for how to normalize

a plane.

[/Notes]

Solution:

We assume is a point inside the frustum. Therefore, its coordinates

in homogeneous space are bounded as

follows:

We can make substitutions to find the plane bounds in world space.

Left Plane:

That is, is in the positive half-space of the world space plane .

Right Plane:


Bottom Plane:


Top Plane:


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Near Plane:


Far Plane:


4. An OBB can be defined by a center point , three orthonormal axis vectors , , and

defining the box orientation, and three extent lengths , , and along the box axes , ,

and , respectivey, that give the distance from the box center to the box sides.

a) Consider Figure 15.13 (which shows the situation in 2D) and conclude the projected “shadow”

of the OBB onto the axis defined by the normal vector is , where

b) In the previous formula for , explain why we must take the absolute values instead of just

computing ?

c) Derive a plane/OBB intersection test that determines if the OBB is in front of the plane,

behind the plane, or intersecting the plane.

d) An AABB is a special case of an OBB, so this test also works for an AABB. However, the

formula for simplifies in the case of an AABB. Find the simplified formula for for the

AABB case.

Solution:

b) If one of the terms is negative, the sum will not give the

“radius” of the OBB. To explain it another way (in 2D), let the vectors from the box center to

the corners be given by:


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[Figure D.13: The corner vectors. Note that these are not the corner points, but the vectors from

the center point to the corner points.]

The OBB “radius” is the corner vector that gives the largest projection onto :

The maximum will be the one where all terms are positive (or all terms are negative) so that

there is no cancellation between the terms. In other words, the maximum will be equal to:

The same argument generalizes to 3D where there are 8 corner vectors.

c) The signed distance from the center of the OBB to the plane is . If then

the sphere intersects the plane. If then the OBB is behind the plane. If then the

OBB is in front of the plane and the sphere intersects the positive half-space of the plane.

d) In the case of an AABB, , , and ; therefore:

Chapter 18


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3. See “DisplacementMappedWaves” on the DVD.

Chapter 21

6. Derive the matrix that maps the box . This is

an “off center” orthographic view volume (i.e., the box is not centered about the view space

origin). In contrast, the orthographic projection matrix derived in §21.2 is an “on center”

orthographic view volume.

Solution:

For all three coordinates, we need to remap an interval onto another interval. We can solve this

problem generally once, and then apply it to each coordinate. We want to map . We assume the mapping takes the form (i.e., a scaling and translation). We have

the conditions and , which allow us to solve for and :

The first equation implies . Plugging this into the second equation we get:

And so:

Therefore,

Applying this formula to our specific intervals, we obtain the transformations:


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Or in terms of matrices:

7. In Chapter 16 we learned about picking with a perspective projection matrix. Derive picking

formulas for an off-centered orthographic projection.

Solution:

Let be the picked point in screen space. Inverting the viewport transformation, we get

the corresponding point in NDC space:

The orthographic projection matrix transforms the view volume from view space to NDC space.

The off-centered orthographic projection transformation is:


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In particular, this gives the two equations:

We can then solve for the view space coordinates in terms of :

Then our picking ray in view space is given by . Note that in an

orthographic projection, all the rays are parallel to the z-axis direction .

Sample code:

Ray3 CalcWorldPickRay(XMMATRIX P, Point s, Size viewport)

{

float sx = (float)s.X;

float sy = (float)s.Y;

float w = (float)viewport.Width;

float h = (float)viewport.Height;

float x = (2.0f*sx/w - 1.0f)/P(0,0) - P(3,0)/P(0,0);

float y = (-2.0f*sy/h + 1.0f)/P(1,1) – P(3,1)/P(1,1);

Ray3 pickRay;

pickRay.Origin = XMFLOAT3(x, y, 0.0f);


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pickRay.Direction = XMFLOAT3(0.0f, 0.0f, 1.0f);

return pickRay;

}

Note: This works for an off-centered orthographic projection matrix and a centered one since a

centered one is just a special case of the off-centered one.

Chapter 24

4. Rotate the vector using complex number multiplication.

Solution:

We can verify this by seeing if we get the same answer with a rotation matrix:

6. Let . Show .

Solution:


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7. Let be a matrix. Prove that and if and only if

. That is, if and only if is a rotation matrix. This gives us a way of testing if

a matrix is a rotation matrix.

Solution:

Let

and suppose and .

along with

imply the following equations:

Now,

Therefore,

Both terms are positive; hence, we must have:

and

So our matrix has the form:


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Finally, because there exists a such that and , and

we get our result:

Now suppose

. Then

Therefore, and . In other words, all rotation matrices have determinant 1

and are orthogonal, and every orthogonal matrix with determinant 1 must be a rotation matrix.

8. Let and be quaternions. Perform the indicated quaternion

operations.

a.

b.

c.

d. e. f.

g.

h.

i.

j.

Solution:

a. b.

c.

d. e.


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f.

g.

h.

i.

j.

XMVECTOR p = XMVectorSet(1.0f, 2.0f, 3.0f, 4.0f);

XMVECTOR q = XMVectorSet(2.0f, -1.0f, 1.0f, -2.0f);

XMVECTOR a = p + q;

XMVECTOR b = p - q;

XMVECTOR c = XMQuaternionMultiply(q,p );

XMVECTOR d = XMQuaternionConjugate(p);

XMVECTOR e = XMQuaternionConjugate(q);

XMVECTOR f = XMQuaternionMultiply(XMQuaternionConjugate(p),p);

XMVECTOR g = XMQuaternionLength(p);

XMVECTOR h = XMQuaternionLength(q);

XMVECTOR i = XMQuaternionInverse(p);

XMVECTOR j = XMQuaternionInverse(q);

cout << "p + q = " << a << endl;

cout << "p - q = " << b << endl;

cout << "pq = " << c << endl;

cout << "p* = " << d << endl;

cout << "q* = " << e << endl;

cout << "p*p = " << f << endl;

cout << "||p|| = " << g << endl;

cout << "||q|| = " << h << endl;

cout << "invP = " << i << endl;

cout << "invQ = " << j << endl;

10. Write the unit quaternion

in polar notation.

11. Find the unit quaternion that rotates about the axis .


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The division by 2 is to compensate for the in Equation 24.3 because we want to rotate by the

angle , not .

XMVECTOR p = XMQuaternionRotationAxis(

XMVectorSet(1.0f, 1.0f, 1.0f, 0.0f),

XMConvertToRadians(45.0f));

cout << "p = " << p << endl;

12. Find the unit quaternion that rotates about the axis .

The division by 2 is to compensate for the in Equation 24.3 because we want to rotate by the

angle , not .

XMVECTOR p = XMQuaternionRotationAxis(

XMVectorSet(0.0f, 0.0f, -1.0f, 0.0f),

XMConvertToRadians(60.0f));

cout << "p = " << p << endl;

13. Let

and

. Compute

and verify it is a unit

quaternion.

Solution:

It is easy to verify that . The angle between the quaternions is given by:


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The interpolated quaternion is easily seen to be unit length:

.

Observe that rotates about the axis by an angle

,

that rotates about the axis by an angle

, and that

rotates about the axis by an angle

. With the

interpolation parameter being the midpoint

, this makes sense, as is right in the middle

between and .

XMVECTOR p = XMVectorSet(0.5f, 0.0f, 0.0f, sqrt(3.0)/2.0f);

XMVECTOR q = XMVectorSet(sqrt(3.0)/2.0f, 0.0f, 0.0f, 0.5f);

XMVECTOR r = XMQuaternionSlerp(p, q, 0.5f);

cout << "r = " << r << endl;

15. Prove that

Solution:

Let .

Because the quaternion has zero vector part, we convert it to a real number as discussed in

§24.2.4. Similarly, it can be shown

.

18. Prove the following properties:


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a. b. c. for

d.

Solution:

Let and so that and .

a)

b)

c)

d)

19. Prove

algebraically.

Solution:

The key component is to apply the trig identity:

Now,


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Appendix C

1. Let be a ray relative to some coordinate system. Plot the points on the

ray at t = 0.0, 0.5, 1.0, 2.0, and 5.0.

Solution:

[FigD.14: Plotting points on a line]

3. For each part, find the vector line equation of the line passing through the two points.

a) b)

Solution:

a)


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b)

5. Let be a line. Find the distance from the following points to the

line:

Solution:

Apply the formula from the previous exercise.


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7. Let

be a plane. Define the locality of the following points relative to the

plane: , , and .

Solution:

The plane equation is:

. Plugging these points into the left-hand side of the

equation gives:

9. Let

be a plane. Find the reflection of the point about the plane.

Solution:

From the previous exercise, we know that is a point on the plane.

Because

, we have that

Now we can apply the formula from §C.4.9:


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10. Let

be a plane, and let be a ray. Find the

point at which the ray intersects the plane. Then write a short program using the

XMPlaneIntersectLine function to verify your answer.

Solution:

We plug into the plane equation to verify it indeed lies on the plane:

The code is given by:

#include <windows.h> // for FLOAT definition

#include <xnamath.h>

#include <iostream>

using namespace std;

// Overload the "<<" operators so that we can use cout to

// output XMVECTOR objects.

ostream& operator<<(ostream& os, FXMVECTOR v)

{

XMFLOAT4 dest;

XMStoreFloat4(&dest, v);

os << "(" << dest.x << ", " << dest.y << ", " << dest.z <<

", " << dest.w << ")";

return os;

}

int main()


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{

XMVECTOR p0 = XMVectorSet(-1.0f, 1.0f, -1.0f, 1.0f);

XMVECTOR u = XMVectorSet(1.0f, 0.0f, 0.0f, 0.0f);

// Construct plane by specifying its (A, B, C, D)

// components directly.

float s = 1.0f / sqrtf(3);

XMVECTOR plane = XMVectorSet(s, s, s, -5.0f);

// Function expects a line segment and not a ray; so we just

// truncate our ray at p0 + 100*u to make a line segment.

XMVECTOR isect = XMPlaneIntersectLine(plane, p0, p0 + 100*u);

cout << isect << endl;

return 0;

}

The output is (in homogeneous coordinates so for points):

(8.66025, 1, -1, 1)

Press any key to continue . . .

We note , so the computer result agrees with our calculation.

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Appendix D Selected Solutions - d3dcoder.net · Solution: a) ... of Exercise 14, we can conclude: ... Appendix D Selected Solutions, Introduction to 3D Game Programming with DirectX